1. A 10.00 mL aliquot of 0.010 M CaCO3 is titrated with 18.22 mL of EDTA...
1. A 10.00 mL aliquot of 0.010 M CaCO3 is titrated with 18.22 mL of EDTA solution. What is the EDTA molarity?
2. A 250.0 mL water sample requires 30.85 mL of the EDTA solution from question 1 to reach the calmagite endpoint. What was the molarity of the hard metal ions in the water sample?
3. If the metal ions in the water sample of question 2 are assumed to be Ca2+ from CaCO3, express the concentration in ppm CaCO3. This is the same as the concentration in units of mg CaCO3 per liter of sample.
Homework Answers
1)
moles of CaCO3 = 10 x 0.01 / 1000
= 1.0 x 10^-4 mol
moles of EDTA = moles of CaCO3
moles of EDTA = 1.0 x 10^-4 mol
moles = molarity x volume
1.0 x 10^-4 = molarity x 0.01822
Molarity of EDTA = 5.49 x 10^-3 M
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